Question

In a convex hexagon, two diagonals are drawn at random. The probability that the diagonals intersect at an interior point of the hexagon, is
(A) $\dfrac{5}{{12}}$
(B) $\dfrac{7}{{12}}$
(C) $\dfrac{2}{5}$
(D) None of these

Answer 1

As we know in the convex hexagon there are 6 points and rest are diagonals. So, we can calculate the total number of diagonals out of which two are intersecting each other. We can easily calculate the probability using this.

Formula used: Here we can use the formula of combination $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$

Complete step-by-step answer:
In a convex hexagon there are 6 points.
So, the total number of ways in which we can join any two vertices in a convex hexagon ${}^6{C_2}$​ .

Out of which there are 6 edges and rest are diagonals.
We can calculate no of diagonals using ${}^6{C_2} - 6$
Here we also use the combination formula $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$ to calculate ${}^6{C_2}$
=$\dfrac{{6!}}{{2!*\left( {6 - 2} \right)!}} - 6$
On simplifying we get,
=$\dfrac{{6!}}{{2!*4!}} - 6$
On solving factorials we get,
= $\dfrac{{6 * 5 * 4!}}{{2*4!}} - 6$
Cancelling 4! From both numerator and denominator.
=$\dfrac{{6 * 5}}{2} - 6$
On further solving we get,
= $\dfrac{{30}}{2} - 6$
On dividing 30 with 2 we get 15.
=$15 - 6 = 9$
So, the total number of diagonals is 9.
As we calculated total number of diagonals that is 9 from which only 2 diagonals can be chosen in ${}^9{C_2}$ using the combination formula $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$ we can calculate ${}^9{C_2}$.
Here, n=9 and r=2
Putting all these values in the above formula we get,
$\Rightarrow$ $^9{C_2} = \dfrac{{9!}}{{\left( {9 - 2} \right)!2!}}$
$\Rightarrow$ $^9{C_2} = \dfrac{{9!}}{{7!2!}}$
Now, we will calculate the factorials,
$\Rightarrow$ $^9{C_2} = \dfrac{{9 * 8 * 7!}}{{7!2!}}$
Cancelling 7! From both numerator and denominator.
We get, $^9{C_2} = \dfrac{{9 * 8}}{2}$
$\Rightarrow$ $^9{C_2} = \dfrac{{72}}{2}$
On dividing 72 by 2 we get:
$\Rightarrow$ $^9{C_2} = 36$ ways
Intersection will occur only if two diagonal intersect that means 4 points can only join.
Therefore, Total number of intersections are ${}^6{C_4}$​ which can be calculated using the combination formula $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$ we can calculate ${}^6{C_4}$.
Here, n=6 and r=4
Putting all these values in the above formula we get,
$\Rightarrow$ $^6{C_4} = \dfrac{{6!}}{{\left( {6 - 4} \right)!4!}}$
$\Rightarrow$ $^6{C_4} = \dfrac{{6!}}{{2!4!}}$
Now, we will calculate the factorials,
$\Rightarrow$ $^6{C_4} = \dfrac{{6 * 5 * 4!}}{{2!4!}}$
Cancelling 4! From both numerator and denominator.
We get, $^6{C_4} = \dfrac{{6 * 5}}{2}$
$\Rightarrow$ $^6{C_4} = \dfrac{{30}}{2}$
On dividing 30 by 2 we get:
$\Rightarrow$ $^6{C_4} = 15$
Hence, the probability that diagonal intersect at an interior point is $\dfrac{{15}}{{36}} = \dfrac{5}{{12}}$

Note: We can solve these types of questions easily with the help of combination formulas. As in this question we just needed to calculate the total number of intersections and the total number of diagonals out of which two are intersecting.